OWAO 2022, Теория — Задача 4. Red Skies

• (B − V)′ = 1.0, (U − B)′ = 0.65; the color excesses of the two components are equal.

Let $V$ be the true (extra-atmospheric) magnitude of each component in the V band. Then we obtain true magnitudes of each component in the U and B bands.

Total magnitude $m_t$ can be calculated from the magnitudes of components $m_1$ and $m_2$ as follows (each light source is compared here to $0^{\mathrm{m}}$):

Then consider $m_1 = V + a$, $m_2 = V + b$:

After the calculation, it is possible to determine the true color indices of the binary star:

Please note that “total” color indices lay between the corresponding color indices of components.

Optical depth $\tau \propto \sigma L$, where $\sigma$ is scattering cross-section and $L$ is beam path length. Recalling Beer’s law,

and in view of the logarithmic nature of stellar magnitudes, we conclude that the change in magnitude due to atmospheric absorption

At an altitude of $45^{\circ}$, we estimate the path of the beam in the atmosphere as

Let us calculate $\Delta m$ for different bands, taking the mean band wavelength as an estimate:

Therefore, after passing through the atmosphere, the color indices of the binary star are

Color excess is a characteristic of the absorbing medium, not of a particular radiation source, so the color excesses of the components match.

Marking Scheme:

• True colors of the binary:

  • Correct formula for the color — 2 pt,
  • Correct values of the colors — 0.5 pt + 0.5 pt.
• $\approx\sqrt{2}$-fold increase of the absorption at an altitude of $45^{\circ}$ — 1 pt.
• Relation between absorption and wavelength — 2 pt.
• Proof of equality of color excesses:

  • Full justification — 1 pt,
  • Correct answer — 1 pt.
• Color excesses estimation — 0.5 pt + 0.5 pt.
• Colors after passing through the atmosphere — 0.5 pt + 0.5 pt.